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Complex Numbers
6 Complex Numbers "In the real world, complex numbers don't exist." Dr. Phillip Christie
6.1 THE NEED FOR COMPLEX NUMBERS All engineers need to know about complex numbers (which they sometimes refer to as j notation), why they are important, and how to handle them. If you would like to see one example of why, you might like to have a quick first look at Section 6.10, before reading each section and, most importantly, doing the exercises. 6.2 THEj NOTATION AND COMPLEX NUMBERS The idea of a complex number springs from the seemingly peculiar step of seeking the solution to the equation (6.1)
Clearly this has no solution in the usual sense, since squaring any "real" number, positive or negative, leads to a positive answer. You might think that is that, but mathematicians made a bold step by supposing there is a solution and calling it i. Engineers call it j so as not to mix it up with electric current, so we shall use that, but be aware when using technology that mathematical software often uses i. This seems to be a strange step to take but, surprisingly, what follows has real meaning and a deep impact in engineering, as you will see later. In detail then j is defined as:
j=~
so that squaring this
J.2 =- 1 .
Complex Numbers
66
[Ch.6
Thusj is a solution to equation (6.1) as required. Becausej is not a number in the way you normally understand, it is necessary to build an understanding of how this bit of mathematics behaves by looking at what happens in various circumstances. Begin by looking at what happens if you repeatedly multiply by j . .3
.. 2
J = J.J j4
= -J..
=jl =j.(-j)=-P =-(-1)=1.
Starting to multiply by j again just goes around the same cycle again: j, -1, -j, 1 and so on. To investigate these ideas further, it is useful to solve more quadratic equations. Example 6.2.1
z 2 =-4 .
Solve
The solution (by hand - you may also try this on your technology) is
z=±.J=4=±.J-lx4=±H.J4=±2j. You see that you do not need any additional strange steps: square roots of any negative number can be expressed in terms ofj in this way. Example 6.2.2
z2 + 4z + 5 = O.
Solve
The solution (using the quadratic formula) is
z=
_4±~42 -4xlx5 -4±.J"=i -4±~.J4 2xl
=
2
=
2
= -4±j2 2+"1 2+' 2 =- -J =- -J.
Once more the only strange step needed is to use j. Again you may try to get your technology to do this: here it is on one graphic calculator: • cSo1ve(z2 + 4'z + 5 = e,z) z= -2+i or z= -2-i
'4&.iN,.,. . . . .·.] MAIN
BAD AUTD
FUNe 1130
Figure 6.1 Solving a quadratic equation on a TI-89
(You will meet more of how these ideas interact with quadratic equations in Chapter 16 when you tackle differential equations.) Note that the answers always come out in the form z = x + jy = x + Y.i ; that is, one part not multiplied by j (x - the real part), and another part multiplied by j (y the imaginary part). A number such as z is called a complex number. The real part of z is defined as Re(z) = x = part not multiplied by j. The imaginary part of z is Im(z) = y = part multiplied by j (but not including j).
67
Complex Arithmetic
Sec. 6.3]
Thus for example Re(3 + j2) = 3 and Im(3 + j2) = 2 (note j is not included). A complex number for which y = 0 is called purely real; one for which x =0 is called purely imaginary. You should see that only one new step is needed - once you are allowed to use j = ~ , all these mathematical problems can be solved in terms of j. No other new steps are needed, although how you think about the new ideas is important.
6.3 ARITHMETIC WITH COMPLEX NUMBERS All the normal rules of arithmetic and algebra apply. Just remember that
J.2 =- 1 . Many calculators and computer packages will handle complex number arithmetic straightforwardly and you should find out about what your technology does. To see how it works, it is best to look at some examples. Suppose then that there are two complex numbers z = 1 + j2 and w = 2 - jl = 2- j. Then addition and subtraction are straightforward: z + w = (1+ j2) + (2 - jl) = (1+ 2) + j(2 -1) = 3 + jl = 3 - j z - w = (1+ j2) - (2 - jl) = 1+ j2 - 2 + jl = (1- 2) + j(2 + 1) = -1 + j3 . Multiplication is also straightforward if you remember about multiplying brackets out (see Chapter 2) and recall that / = -1 . z.w=(1 + j2).(2 - jl) = lx2+ lx(-jl) +(j2)x2+ (j2)x (-jl) = 2 - jl + j4
-l.2 = 2 -
jl + j4 - (-1).2 = 4 + j3
Division is slightly more complicated by hand (although not with your technology of course!) because you want to write the answer in the form x + jy, so you need an algebraic trick to get the j off the bottom of the fraction. This trick involves the complex conjugate. The complex conjugate of a complex number z = x + jy is defined as Z or z* = x - jy (that is, the conjugate is the same as Z except that the imaginary part has had its sign changed). It has the special property that for any complex number,
z.z = (x + jy)(x -
2
jy) = x + /
2
+ OJ = x + /
which is purely real.
Now to sort out division of complex numbers, you just multiply the top and the bottom of your quotient by the complex conjugate of the bottom. Thus for example: w
z
(2 - jl) (1+ j2)
=---=
(2 - jl)(1- j2) (1 + j2)(l- j2)
(multiplying top and bottom by conjugate of bottom),
68
Complex Numbers (2- jl- j4+ j (0- p) = 24) (l+j2-j2-j (5) 22)
=
[Ch.6
0'1 . - J =-J.
Note we have tidied up and taken care with signs; j has disappeared from the bottom as planned, and in this example the real part happens to be 0 so at the end we leave it out. Checking on the graphic calculator:
Figure 6.2 Complex arithmetic on a 11-89
DO THIS NOW! 6.1 Complex arithmetic on calculators: Most recent calculators will handle complex arithmetic. Find out now how your technology handles complex numbers, concentrating for now on how to add, subtract, multiply, raise to a power and divide. Identify whether your technology uses "i" or 'T' or some other form; check that you know how complex arithmetic works by doing the examples below. Work out how to find conjugates and real and imaginary parts. Talk to someone about it if you cannot sort it out.
Solve aU of the following problems, both by hand and using appropriate technology, and thus check your own answers! 1. Solve the quadratic equations
(a) x 2
=-9
(b) x + 2x + 2 = 0 . 2
In each case identify the real and imaginary parts of the roots. 2. Let
z =1+ j and w = -2- 5j . Calculate by hand and check by technology:
(a)
z+w
(b)
z-w
(c)
z,w
(d)
z
(e)
w
(1)
z.Z
(g)
2w-3z
(h)
w/z
(i)
Re(z.w)
(j)
Im(z,w)
6.4 GEOMETRY WITH COMPLEX NUMBERS: THE ARGAND DIAGRAM It is always useful to have a way of thinking about a mathematical idea through pictures. With complex numbers, the geometrical model for visualisation is called an Argand diagram. In this, the complex number z = x + jy is represented by the point (x, y) in a two-dimensional plane, as in Figure 6.3.
69
The Argand Diagram
Sec. 6.4]
Figure 6.3 An Argand Diagram
The horizontal axis is called the real axis, as this measures x, the real part of z, The vertical axis is called the imaginary axis, as this measures y, the imaginary part of z, The complex conjugate of z, z* or Z • appears as the reflection of z in the real axis. A complex number then is in a sense a two-dimensional quantity, represented by two real numbers in a certain order, x first, then y, i.e. an orderedpair. Indeed some graphic calculators reflect this by showing the complex number x + jy as (x, y).
ICxample6.4.1 Let z =1+ j and w = -2 - 3j . Observe the positions on the Argand diagram in Figure 6.4. lm
z Re
z* w
Figure6.4 An exampleof an Arganddiagram
DO TIllS NOW! 6.2 Plot these complex numbers on an Argand diagram: (i) 1+ OJ(= I)
(ii) O+Ij(=j)
(iv) o-Ij(=l)
(v) I+Oj(=j4)
(iii) -I +OJ (=
f)
What do you observe about the five numbers you have plotted? How would you describe what is done to the position of a complex number on the Argand diagram when you multiply by j? Think about and answer this before reading on into the following bracket.
(You should get the idea that multiplying by j rotates the complex number anticlockwise through 90 ~ Try it with some other examples ofyour own now.)
70
[Ch.6
Complex Numbers
Dealing with a geometrical representation here leads to ideas for other ways of representing complex numbers and this is followed up in the next section. 6.5 CARTESIAN AND POLAR FORM, MODULUS AND ARGUMENT. The Argand diagram makes it clear that to each complex number there corresponds a unique point on the x-y plane. Thus any way we can define how to reach a point on that plane gives another way of representing the complex number. This leads to the idea of polar form.
Figure 6.5 Polar form and the Argand diagram
Look at Figure 6.5. The standard definition of x + jy says "start at the origin, go x units in the real (Re) direction, then go y units in the imaginary (1m) direction". This is called the Cartesian or rectangular form of the complex number. An alternative way of reaching that same point is to say "start at the origin facing in the real direction, turn through an angle 0 anti-clockwise, then go r units in that direction. This is known as the polar form of the complex number, sometimes written rLO, (r,O) or rcisO. Since these two forms represent the same number (and point), then there must be a way of going from one form to the other; this comes from the right angled triangle.
Figure 6.6 Geometrical connection between Cartesian and polar form
From Figure 6.6, two relationships are apparent, using either Pythagoras' theorem, or the basic definitions of the trigonometric functions: Converting from (r, 8) to (x, y) x rcos(O)
=
y = rsin(O)
Converting from (x, y) to (r, 8)
r
= ~rx-2-+-y-2
o = arctan( ~ ) Beware the multiple values of the arctan function though - see section 4.6.1 and Example 6.5 below.
Sec. 6.5]
71
Polar Form of Complex Numbers
There is some jargon associated with this:
r= (J
~x2 + i
is the modulus of z, which can also be written
=arctan( ~ )
Izi or abs(z).
is the argument of z, which can also be written arg z or angle(z)
= r sin (J , then by replacing x and y in z = x + jy is a different way of writing z, as z = r cos (J + jr sin (J = r(cos (J + j sin (J) . Since x
= r cos (J
and y
Summary: z
there
= x + jy = r(cos (J + j sin (J) = rL (J
Example 6.5.1 Convert the complex number z = 1 + jl to polar form. Applying the formulae:
r=~ =/2 z1.414. (J
f)
= arctan( = arctan(l) = 45° =:
radians z 0.785 radians.
i'n
~-4- ..f2
• 1+i
.1 + i
'UI MAIN
~.785398·i·1.41421
rUNe sno
RAD AUTO
Figures 6.7(a) and (b) Checking conversion by picture and by technology
Figures 6.7(a) and (b) shows how this ties in geometrically, and with the technology. Note the currently unfamiliar way this particular machine presents the result. See Section 6.6 for some explanation of this. Check the conversion using any direct P-R or R-P converter on your own technology. Example 6.5.2 There is a complication with the fact that the tangent function repeats every 180 ~ so when using the arctan function you should also check that you are in the correct quadrant, as in this example. Convert the complex number z = -1- jl to polar form. Applying the formulae:
r = ~(_1)2 +(_1)2 (J =
(-1)=
arctan -
-1
= /2 "" 1.414.
arctan(l) = 45° =-~ rads
4
=????
Now on the face of it this looks like it is turning out exactly the same as Example 6.5.1, but that cannot be right since they are different numbers in different positions on the Argand diagram
72
[Ch.6
Complex Numbers
To sort out what is happening, look at Figure 6.8:
Figure 6.8 Watch your quadrant! o
You should note that tan(45 )
=1, but so also does tan(-135 ) =1 (check on your 0
calculator). Thus arctan(l) has two possible values, 45° and -135°, separated by 180°. To see which is the right one ALWAYS DRAW THE PICTURE TO CHECK. As a back-up, you should see that the direct P-R or R-P converters or equivalent on your technology will sort this out automatically. In this case then the answer is shown and confirmed by technology in Figure 6.9.
,= 1= (-=1) = 1.414
8
arctan
-1
arctan( 1) = -135
0
3Hc
=--4-
Figure 6.9 Checking the answer using technology - see Section 6.6 for explanation of notation here.
Example 6.5.3 Convert 5L2.2 C to rectangular form.
5L2.2
c
=5(cos(2.2) + j sin(2.2»
... -2.94 + j4.04.
The multi-value complication does not occur this way round. DO TIDS NOW! 6.3 In this box, Z = 2 + j5 and w = -1 - j2 . In each of the following problems, do the calculations by hand and check by using the direct P-R or R-P converters or your own equivalent technology, and also by drawing Argand diagrams. 1. Evaluate: (i) Izi (ii) arg(z). 2. Convert both z and w to polar form, either using the formulae, or using your technology, or preferably both, quoting the angles in both degrees and radians. Confirm that your solutions are correct by placing them on an Argand diagram and seeing if the diagram fits with your figures.
3. Convert to rectangular form (a) 2LI.3 C (b) 3L35° .
Euler's Relationship and Exponential Form
Sec. 6.6]
73
6.6 EULER'S RELATIONSHIP AND EXPONENTIAL FORM One of the most astounding properties of complex numbers is Euler's relationship. This shows that, if you raise e to a complex power, then the exponential function is very closely related to trigonometric functions. This may surprise you, as you may be used to exponentials representing decay or growth, and sines and cosines representing oscillations. Nevertheless, without any proof, here it is - put it in your personal handbook. Euler's relationship:
e
j8
=cosO+ jsinO
(Note 0 must be in radians)
This amazing result is surprisingly very useful in engineering mathematics, especially in analysing communication systems, and solving differential equations. You will meet the justification of it in Chapter 18, but for now it is useful to become familiar with it, so here are one or two examples.
Example 6.6.1
e
j~ =cos(K) . . (K) .J2 07071 + J'0.7071. - + Jsm - =J2 - + J-::::. 4
4
Example 6.6.2 Here is a very nice one: e
2
j Jr
2
= (cos K + j sin K) = (-1 + jO) = -1 .
In other words, combine s; which you can never know exactly, with e which you can never know exactly, with j, which is a figment of your imagination - and you end up with just the number -1 . This is a remarkable result!
Exponential form What Euler's relationship gives you is another very useful way of representing complex numbers. You have met already: z = x + jy = r(cos 0 + j sin 8) But from Euler's relationship, cosO + j sin 0 = e the form
je
, and so you can write z in
z = x + jy = r( cos 0 + j sin 0) = re j8
-
remember 0 in radians!
This is called the exponential form of the complex number z. You will note that it does not require any more calculation to find it. Once you have polar form then you have exponential form, as long as () is in radians. This might help you to understand the notation appearing in the technology screens in Figures 6.7 and 6.9.
Example 6.6.3
z = 1+ jl "" 1.414(cos 45° + jsin 45°) from Example 6.5.1 "" 1.414(cos ll' + j sin ll' ) = 1.414L ll' 4 4 4 .Jr
= 1,414e
J4
with the angle in radians.
Complex Numbers
74
[Ch.6
DO TIllS NOW! 6.4 1. You have already converted Z = 2 + j5 and w = -1 - j2 to polar form in "DO THIS NOW 6.3!". Convert z and w to exponential form.
2. Euler's relationship allows you to find more complicated complex exponentials. If z =x + jy, the result allows you to find jy x jy e' =e x+ =e e =eX(cosy+jsiny).
Evaluate the following (by hand and/or by technology - how directly will your machine do this? Find out NOW): (a)
ej
(b)
e-1.I+3j
(c)
Re(e-1.I+3 j )
(d)
Im(e-1.1+3j).
3. If you plot Re(e(-1.1+3j)x) you are plotting the graph of y
= e-l.1x cos(3x) .
Satisfy yourself about this both algebraically and by plotting it on your technology. 4. Experiment with curves of the type
y
=e
QX
cos(bx)
and y = e sin(bx) , as suggested below. QX
Describe in words what happens when a is (a) positive, (b) 0, (c) negative. Describe what happens when b is changed from small to large values. Why do you think this type of function is useful in engineering? What kind of motion can it represent? 5. If
z = -0.1 + j3
and t is an independent variable, use Euler's relationship to
write down (b)
Im(e
2zt)
Predict the shape of the graphs of both these functions for t ~ 0 and confirm by plotting them on your technology. Explain what you see in words. If you use your technology to turn symbols into pictures, you can "see" the importance and use of Euler's relationship in engineering. Figures 6.10(a) and (b) show an example of how a complex exponential can represent damped oscillations:
Figures 6.1O(a) and (b) Complex exponentials represent damped oscillations
Some uses of Polar and Exponential Form
Sec. 6.7]
75
6.7 SOME USES OF POLAR AND EXPONENTIAL FORM It turns out that it is particularly neat to handle multiplication, division and powers of complex numbers in polar form. Here is why. Have a look at these two complex numbers: ZI
'8
= rlLBI = 'ieJ
I
'(J
and Z2 = r 2LB2 = r 2e J 2.
Then their product is Gust using the laws of indices) ZIZ2
= ('iLBl)(r2LB2) = ('iej~ )(r2e
j(J2)
= 'ir2ej(~+(J2) = 'i r2L(BI +(2 ) .
I To multiply two complex numbers, multiply the moduli and add the arguments. Their quotient is (again just using the laws of indices)
.s., (rIL81) = (rle. 8) =.2-ej(~-82) =.2-L(8 j~
Z2
(r2 L B2)
(r2eJ 2)
r2
1
r2
-( 2).
I To divide two complex numbers, divide the moduli and subtract the arguments. To raise to a power n (once more the laws of indices are needed):
To raise a complex number to the power n, raise the modulus to the power n and multiply the argument by n, Example 6.7.1 Let z = 3 - 2j and w = 1- j. These give (check on your technology) in polar form c c 1r Z "" (3.606LO.588 ) and w "" (1.414L - 4" ) . Plotting these on an Argand diagram together with jz,
pz , j3z and z.w
gives Figure 6.11:
---
zw=(5.1<-.2D]
).).z
1·I·j·z
Figure 6.11 The picture shows the same story as the algebra
76
Complex Numbers
[Ch.6
The multiplication by powers of j again illustrates the rotational aspect of j (900
in the anticlockwise direction) and zw shows the addition of angles 0.588 + (- f) and multiplication of the moduli (3.606*1.414).
DO THESE NOW! 6.5 1. Convert the numbers 2 + 3j and 4 - 3j to polar form. HENCE evaluate the following: ..) (11
(i) (2+3j)(4-3j)
2+3'J 4-3j
(iii)
--
2
( J 2 + 3j 4-3j
Check your answers with your technology (Beware! Make sure you know and understand in what form they appear.) Show the two original numbers on an Argand diagram. Add to the diagram your answers to (i), (ii) and (iii), explaining how the rules in polar form, particularly concerning adding and subtracting angles, show themselves on the diagram. 2. Evaluate (a) (2+3j)3
(b)
(0.1-0.2j)9
Check your answers here by using technology.
6.8 COMPLEX ALGEBRA All of the above principles apply when you deal with complex algebra as opposed to just arithmetic, that is, with letters instead of or as well as numbers. The main differences are (a) you will have to do some algebra (!), and (b) you will need to use a CAS technology to check answers or to work out the more complicated cases. Example 6.8.1 Figures 6.12(a) and (b) show a few simple algebraic examples, carried out on handheld CAS technology, illustrating some basic rules. Can you identify them?
-liPrgPlIOlTeClean
J(
f l · III n · f~· •l:"11l!lf ~ Algebra Calc Other .~a + b'i)'~c + d'i)
f'·
Up,T
II:l~lAI ;~bralc~lclO{~;r IPr~Pl I 0 be~'; Up1
a'c - b·d +(a'd + b'c)'i • tCollect« rl .! al)'< r2 .! a2» -a',f2 a·,f2 . .,i '(al + a2)'rl'r2 2 T + ""2-'b. & .,i '(al - a2)'rl • tC 11 t« rl .! a1)) o ec
•
;x)
.<2.!45O) < 3 .!
-x)
ttll"'*"1ffJJ MAIN
fS
AD AurD
.,
fUNe )/99
'x
_'dliC"riti.Wil MAIN
UD Aura
fUNC )/'
Figures 6.12(a) and (b) Complex algebra on a hand-held CAS system
Sec. 6.8]
Complex Algebra
77
DO THIS NOW! 6.6 Find out NOW how your CAS handles complex numbers. Test it out on the examples below, which as usual you should do on paper as well as on your machine. I. Find the real and imaginary parts of z when I 2 1 -=--+-z 3+2j 4-3j 2. If z =a + jb and w = c + jd confirm by paper and pencil that your CAS is giving appropriate answers to the following: z+w
Izi
zw
z-w 3z-2w
zz
z w
1
arg(w)
(w+z)
You may also like to substitute some numerical values in for a, b, c and d for checking purposes. Make sure you know about and can use the different modes of CAS output (rectangular, polar or exponential forms) available to you. 6.9 ROOTS OF COMPLEX NUMBERS De Moivre'sTheorem
r
Note that (cos(B) + jsin(B))n = (e j 9 = e j n9 = cos(nB) + jsin(nB). This result is known as de Moivre's theorem. This helps to clarify and inform the way to raise a complex number to a fractional power - for instance to take square roots: zn
= r" e j n9 = r" (cos(nB) + j sin(nB)) .
In fact it is a little more complicated than this, since sine and cosine have period 21t, so increasing Bby 21t will change nothing. Thus in general h k'IS any In . t eger (0 ,_ +1,_+2,_, + 3....... ) ,and so z = re j9 = re j(9+2JZk) were
Normally the possibility of an extra 21t does not matter, but it does when n is a fraction, as it will indeed be when finding roots of complex numbers. Example 6.9.1 Solve the complex equation
z 2 = -9 .
In general exponential form
z2
=-9 =ge j(tr+2JZk)
Taking the square roots of both sides gives ..!.
z = (-9) 2
..!.
=9 2 e j (tr+2JZk )12 = 3e j (Jr+2JZk )/ 2 = 3e j (Jr /2+JZk)
Note that we take the positive square root of 9 here.
78
Complex Numbers
[Ch.6
To interpret this, remember that k can take any integer values. All values of z have modulus 3 - that is, they lie on the circle centred on the origin of radius 3. Now at what angles do they lie? Taking k
= 0 gives a z value at () =~ . Taking k = 1
gives a z value at () = 3~ . Taking further values of k simply repeats these two positions. There are thus just the two distinct values and just two square roots of -9. A diagram is a good check here since all the roots have the same modulus and in this case have an angle of 21t12 between their arguments.
fm 21
No_2p1l2 ~.ba
arc-.. •f . .t ....... a
1
2.2
Re
2
Figure 6.13 Roots are equally spaced around the circle ~lUlmple6.9.2
Solve the equation Z3
=0.1- 0.2j.
In exponential form we have 0.1- 0.2j given by
0.606962e j (0.36905+211k / 3)
i.e. 0.606962e j (O.36905)
,
= .223607 e j (1.10715 + 211k)
so all solutions are
with k = 0, 1 and 2.
0.606962e j (2.46345) , 0.606962e j (4.455784) •
Note that in this case the roots are all separated by 21t13 (120°) and effectively cut = 0.606962 into 3 equal sections. the circle
Izl
6.10 MINI CAS~ STUDY Complex numbers occur, for instance, in two important areas of engineering: electronic circuit theory and differential equation theory. Because you meet more about differential equations in Chapter 16 onwards, we choose to present here a little case study about an electronic circuit. An alternating current flowing in the circuit shown in Figure 6.14 is given by I = 10 sin(aJt) (OJ is a constant and t is time). The corresponding voltage drops depend on OJ and on the resistance R, capacitance C and inductance L of the circuit.
Sec. 6.10]
79
Complex Numbers: Exercises
For a resistance R, the voltage V is V = loR sin(ox) and this is "in phase" with the current. R L
c Figure 6.14 A simple series LCR circuit
For a capacitance C, the voltage V
=!JL sineOJ t c£
1! ) , a phase "lag" of
2
the current.
!!.2
behind
For an inductance L, the voltage V = OJ L 10 sin(OJt + 1!), a phase "lead" of
2
ahead of the current.
1r
2
If the current I = 10 sin(wt) is regarded as the imaginary part of a complex number
I
= Im(loe jOX ) then these various voltage drops can be represented very
succinctly
by recalling (refer back to Figure 6.11) that multiplication by j gives a lag of division by j (which is the same as multiplication by -j) gives a lead of Thus for resistance
V=RI.
For capacitance
V = -jI .
For inductance
V=jOJLI.
!!.2
and
!!... 2
«c
In general if V = IZ the factor Z is called the complex impedance, which is thus Z= R for a resistor, Z =
_--L for a capacitor, and ox:
Z = jOJL for an inductor.
END OF CHAPTER 6 - MIXED EXERCISES - DO ALL THESE NOW!
For each of the following, predict the answer by hand, check and observe by your technology, and explain and put right any discrepancies. 1. Let z\ = 2 - j, Zz = -1 + 2j, Z3 = 3 + 4j . Evaluate without using polar form, and present your answers in Cartesian form: (ii)
80
Complex Numbers (iii) z\ Z2
[Ch.6
(iv)
Z3
2. Where it is an advantage to do so, repeat the calculations of question 1 using polar form, and present your answers in polar form. 3. If Z =-o.1+5jand t is an independent variable, use Euler's relationship to write down and plot the resulting function of t, for t ~ 0 in each of (i) and (ii): (ii) Im(2e zt )
•
4. The voltage E across an alternating current circuit is obtained by multiplying the total impedance, Z, of the circuit by the current I flowing in the circuit: i.e,
E=I.Z.
A certain circuit, in which a current I = 4 + 5j flows, consists of three separate impedances, Z\, Z2 and Z3 arranged so that the total impedance Z is given by Z = Z\ + (Z2Z3) I(Z2 + Z3)' Obtain the symbolic expression for the voltage E and calculate its value (lEI), given (i)
Z\ = 2+3j
Z2 =4+3j
Z3 =2-5j
(ii)
Z\ = a+bj
Z2=4+3j
Z3 =2-5j.
5.
Solve the equation
6.
(Harder) Given -9+40j
Z3
= 8 + j8 and plot the results on an Argand diagram.
z = 4 + 5 j, write the following numbers in terms of z.
.09756-.12195j
-236-115j
1.7921c
.15617.
6.1 THE NEED FOR COMPLEX NUMBERS All engineers need to know about complex numbers (which they sometimes refer to as j notation), why they are important, and how to handle them. If you would like to see one example of why, you might like to have a quick first look at Section 6.10, before reading each section and, most importantly, doing the exercises. 6.2 THEj NOTATION AND COMPLEX NUMBERS The idea of a complex number springs from the seemingly peculiar step of seeking the solution to the equation (6.1)
Clearly this has no solution in the usual sense, since squaring any "real" number, positive or negative, leads to a positive answer. You might think that is that, but mathematicians made a bold step by supposing there is a solution and calling it i. Engineers call it j so as not to mix it up with electric current, so we shall use that, but be aware when using technology that mathematical software often uses i. This seems to be a strange step to take but, surprisingly, what follows has real meaning and a deep impact in engineering, as you will see later. In detail then j is defined as:
j=~
so that squaring this
J.2 =- 1 .
Complex Numbers
66
[Ch.6
Thusj is a solution to equation (6.1) as required. Becausej is not a number in the way you normally understand, it is necessary to build an understanding of how this bit of mathematics behaves by looking at what happens in various circumstances. Begin by looking at what happens if you repeatedly multiply by j . .3
.. 2
J = J.J j4
= -J..
=jl =j.(-j)=-P =-(-1)=1.
Starting to multiply by j again just goes around the same cycle again: j, -1, -j, 1 and so on. To investigate these ideas further, it is useful to solve more quadratic equations. Example 6.2.1
z 2 =-4 .
Solve
The solution (by hand - you may also try this on your technology) is
z=±.J=4=±.J-lx4=±H.J4=±2j. You see that you do not need any additional strange steps: square roots of any negative number can be expressed in terms ofj in this way. Example 6.2.2
z2 + 4z + 5 = O.
Solve
The solution (using the quadratic formula) is
z=
_4±~42 -4xlx5 -4±.J"=i -4±~.J4 2xl
=
2
=
2
= -4±j2 2+"1 2+' 2 =- -J =- -J.
Once more the only strange step needed is to use j. Again you may try to get your technology to do this: here it is on one graphic calculator: • cSo1ve(z2 + 4'z + 5 = e,z) z= -2+i or z= -2-i
'4&.iN,.,. . . . .·.] MAIN
BAD AUTD
FUNe 1130
Figure 6.1 Solving a quadratic equation on a TI-89
(You will meet more of how these ideas interact with quadratic equations in Chapter 16 when you tackle differential equations.) Note that the answers always come out in the form z = x + jy = x + Y.i ; that is, one part not multiplied by j (x - the real part), and another part multiplied by j (y the imaginary part). A number such as z is called a complex number. The real part of z is defined as Re(z) = x = part not multiplied by j. The imaginary part of z is Im(z) = y = part multiplied by j (but not including j).
67
Complex Arithmetic
Sec. 6.3]
Thus for example Re(3 + j2) = 3 and Im(3 + j2) = 2 (note j is not included). A complex number for which y = 0 is called purely real; one for which x =0 is called purely imaginary. You should see that only one new step is needed - once you are allowed to use j = ~ , all these mathematical problems can be solved in terms of j. No other new steps are needed, although how you think about the new ideas is important.
6.3 ARITHMETIC WITH COMPLEX NUMBERS All the normal rules of arithmetic and algebra apply. Just remember that
J.2 =- 1 . Many calculators and computer packages will handle complex number arithmetic straightforwardly and you should find out about what your technology does. To see how it works, it is best to look at some examples. Suppose then that there are two complex numbers z = 1 + j2 and w = 2 - jl = 2- j. Then addition and subtraction are straightforward: z + w = (1+ j2) + (2 - jl) = (1+ 2) + j(2 -1) = 3 + jl = 3 - j z - w = (1+ j2) - (2 - jl) = 1+ j2 - 2 + jl = (1- 2) + j(2 + 1) = -1 + j3 . Multiplication is also straightforward if you remember about multiplying brackets out (see Chapter 2) and recall that / = -1 . z.w=(1 + j2).(2 - jl) = lx2+ lx(-jl) +(j2)x2+ (j2)x (-jl) = 2 - jl + j4
-l.2 = 2 -
jl + j4 - (-1).2 = 4 + j3
Division is slightly more complicated by hand (although not with your technology of course!) because you want to write the answer in the form x + jy, so you need an algebraic trick to get the j off the bottom of the fraction. This trick involves the complex conjugate. The complex conjugate of a complex number z = x + jy is defined as Z or z* = x - jy (that is, the conjugate is the same as Z except that the imaginary part has had its sign changed). It has the special property that for any complex number,
z.z = (x + jy)(x -
2
jy) = x + /
2
+ OJ = x + /
which is purely real.
Now to sort out division of complex numbers, you just multiply the top and the bottom of your quotient by the complex conjugate of the bottom. Thus for example: w
z
(2 - jl) (1+ j2)
=---=
(2 - jl)(1- j2) (1 + j2)(l- j2)
(multiplying top and bottom by conjugate of bottom),
68
Complex Numbers (2- jl- j4+ j (0- p) = 24) (l+j2-j2-j (5) 22)
=
[Ch.6
0'1 . - J =-J.
Note we have tidied up and taken care with signs; j has disappeared from the bottom as planned, and in this example the real part happens to be 0 so at the end we leave it out. Checking on the graphic calculator:
Figure 6.2 Complex arithmetic on a 11-89
DO THIS NOW! 6.1 Complex arithmetic on calculators: Most recent calculators will handle complex arithmetic. Find out now how your technology handles complex numbers, concentrating for now on how to add, subtract, multiply, raise to a power and divide. Identify whether your technology uses "i" or 'T' or some other form; check that you know how complex arithmetic works by doing the examples below. Work out how to find conjugates and real and imaginary parts. Talk to someone about it if you cannot sort it out.
Solve aU of the following problems, both by hand and using appropriate technology, and thus check your own answers! 1. Solve the quadratic equations
(a) x 2
=-9
(b) x + 2x + 2 = 0 . 2
In each case identify the real and imaginary parts of the roots. 2. Let
z =1+ j and w = -2- 5j . Calculate by hand and check by technology:
(a)
z+w
(b)
z-w
(c)
z,w
(d)
z
(e)
w
(1)
z.Z
(g)
2w-3z
(h)
w/z
(i)
Re(z.w)
(j)
Im(z,w)
6.4 GEOMETRY WITH COMPLEX NUMBERS: THE ARGAND DIAGRAM It is always useful to have a way of thinking about a mathematical idea through pictures. With complex numbers, the geometrical model for visualisation is called an Argand diagram. In this, the complex number z = x + jy is represented by the point (x, y) in a two-dimensional plane, as in Figure 6.3.
69
The Argand Diagram
Sec. 6.4]
Figure 6.3 An Argand Diagram
The horizontal axis is called the real axis, as this measures x, the real part of z, The vertical axis is called the imaginary axis, as this measures y, the imaginary part of z, The complex conjugate of z, z* or Z • appears as the reflection of z in the real axis. A complex number then is in a sense a two-dimensional quantity, represented by two real numbers in a certain order, x first, then y, i.e. an orderedpair. Indeed some graphic calculators reflect this by showing the complex number x + jy as (x, y).
ICxample6.4.1 Let z =1+ j and w = -2 - 3j . Observe the positions on the Argand diagram in Figure 6.4. lm
z Re
z* w
Figure6.4 An exampleof an Arganddiagram
DO TIllS NOW! 6.2 Plot these complex numbers on an Argand diagram: (i) 1+ OJ(= I)
(ii) O+Ij(=j)
(iv) o-Ij(=l)
(v) I+Oj(=j4)
(iii) -I +OJ (=
f)
What do you observe about the five numbers you have plotted? How would you describe what is done to the position of a complex number on the Argand diagram when you multiply by j? Think about and answer this before reading on into the following bracket.
(You should get the idea that multiplying by j rotates the complex number anticlockwise through 90 ~ Try it with some other examples ofyour own now.)
70
[Ch.6
Complex Numbers
Dealing with a geometrical representation here leads to ideas for other ways of representing complex numbers and this is followed up in the next section. 6.5 CARTESIAN AND POLAR FORM, MODULUS AND ARGUMENT. The Argand diagram makes it clear that to each complex number there corresponds a unique point on the x-y plane. Thus any way we can define how to reach a point on that plane gives another way of representing the complex number. This leads to the idea of polar form.
Figure 6.5 Polar form and the Argand diagram
Look at Figure 6.5. The standard definition of x + jy says "start at the origin, go x units in the real (Re) direction, then go y units in the imaginary (1m) direction". This is called the Cartesian or rectangular form of the complex number. An alternative way of reaching that same point is to say "start at the origin facing in the real direction, turn through an angle 0 anti-clockwise, then go r units in that direction. This is known as the polar form of the complex number, sometimes written rLO, (r,O) or rcisO. Since these two forms represent the same number (and point), then there must be a way of going from one form to the other; this comes from the right angled triangle.
Figure 6.6 Geometrical connection between Cartesian and polar form
From Figure 6.6, two relationships are apparent, using either Pythagoras' theorem, or the basic definitions of the trigonometric functions: Converting from (r, 8) to (x, y) x rcos(O)
=
y = rsin(O)
Converting from (x, y) to (r, 8)
r
= ~rx-2-+-y-2
o = arctan( ~ ) Beware the multiple values of the arctan function though - see section 4.6.1 and Example 6.5 below.
Sec. 6.5]
71
Polar Form of Complex Numbers
There is some jargon associated with this:
r= (J
~x2 + i
is the modulus of z, which can also be written
=arctan( ~ )
Izi or abs(z).
is the argument of z, which can also be written arg z or angle(z)
= r sin (J , then by replacing x and y in z = x + jy is a different way of writing z, as z = r cos (J + jr sin (J = r(cos (J + j sin (J) . Since x
= r cos (J
and y
Summary: z
there
= x + jy = r(cos (J + j sin (J) = rL (J
Example 6.5.1 Convert the complex number z = 1 + jl to polar form. Applying the formulae:
r=~ =/2 z1.414. (J
f)
= arctan( = arctan(l) = 45° =:
radians z 0.785 radians.
i'n
~-4- ..f2
• 1+i
.1 + i
'UI MAIN
~.785398·i·1.41421
rUNe sno
RAD AUTO
Figures 6.7(a) and (b) Checking conversion by picture and by technology
Figures 6.7(a) and (b) shows how this ties in geometrically, and with the technology. Note the currently unfamiliar way this particular machine presents the result. See Section 6.6 for some explanation of this. Check the conversion using any direct P-R or R-P converter on your own technology. Example 6.5.2 There is a complication with the fact that the tangent function repeats every 180 ~ so when using the arctan function you should also check that you are in the correct quadrant, as in this example. Convert the complex number z = -1- jl to polar form. Applying the formulae:
r = ~(_1)2 +(_1)2 (J =
(-1)=
arctan -
-1
= /2 "" 1.414.
arctan(l) = 45° =-~ rads
4
=????
Now on the face of it this looks like it is turning out exactly the same as Example 6.5.1, but that cannot be right since they are different numbers in different positions on the Argand diagram
72
[Ch.6
Complex Numbers
To sort out what is happening, look at Figure 6.8:
Figure 6.8 Watch your quadrant! o
You should note that tan(45 )
=1, but so also does tan(-135 ) =1 (check on your 0
calculator). Thus arctan(l) has two possible values, 45° and -135°, separated by 180°. To see which is the right one ALWAYS DRAW THE PICTURE TO CHECK. As a back-up, you should see that the direct P-R or R-P converters or equivalent on your technology will sort this out automatically. In this case then the answer is shown and confirmed by technology in Figure 6.9.
,= 1= (-=1) = 1.414
8
arctan
-1
arctan( 1) = -135
0
3Hc
=--4-
Figure 6.9 Checking the answer using technology - see Section 6.6 for explanation of notation here.
Example 6.5.3 Convert 5L2.2 C to rectangular form.
5L2.2
c
=5(cos(2.2) + j sin(2.2»
... -2.94 + j4.04.
The multi-value complication does not occur this way round. DO TIDS NOW! 6.3 In this box, Z = 2 + j5 and w = -1 - j2 . In each of the following problems, do the calculations by hand and check by using the direct P-R or R-P converters or your own equivalent technology, and also by drawing Argand diagrams. 1. Evaluate: (i) Izi (ii) arg(z). 2. Convert both z and w to polar form, either using the formulae, or using your technology, or preferably both, quoting the angles in both degrees and radians. Confirm that your solutions are correct by placing them on an Argand diagram and seeing if the diagram fits with your figures.
3. Convert to rectangular form (a) 2LI.3 C (b) 3L35° .
Euler's Relationship and Exponential Form
Sec. 6.6]
73
6.6 EULER'S RELATIONSHIP AND EXPONENTIAL FORM One of the most astounding properties of complex numbers is Euler's relationship. This shows that, if you raise e to a complex power, then the exponential function is very closely related to trigonometric functions. This may surprise you, as you may be used to exponentials representing decay or growth, and sines and cosines representing oscillations. Nevertheless, without any proof, here it is - put it in your personal handbook. Euler's relationship:
e
j8
=cosO+ jsinO
(Note 0 must be in radians)
This amazing result is surprisingly very useful in engineering mathematics, especially in analysing communication systems, and solving differential equations. You will meet the justification of it in Chapter 18, but for now it is useful to become familiar with it, so here are one or two examples.
Example 6.6.1
e
j~ =cos(K) . . (K) .J2 07071 + J'0.7071. - + Jsm - =J2 - + J-::::. 4
4
Example 6.6.2 Here is a very nice one: e
2
j Jr
2
= (cos K + j sin K) = (-1 + jO) = -1 .
In other words, combine s; which you can never know exactly, with e which you can never know exactly, with j, which is a figment of your imagination - and you end up with just the number -1 . This is a remarkable result!
Exponential form What Euler's relationship gives you is another very useful way of representing complex numbers. You have met already: z = x + jy = r(cos 0 + j sin 8) But from Euler's relationship, cosO + j sin 0 = e the form
je
, and so you can write z in
z = x + jy = r( cos 0 + j sin 0) = re j8
-
remember 0 in radians!
This is called the exponential form of the complex number z. You will note that it does not require any more calculation to find it. Once you have polar form then you have exponential form, as long as () is in radians. This might help you to understand the notation appearing in the technology screens in Figures 6.7 and 6.9.
Example 6.6.3
z = 1+ jl "" 1.414(cos 45° + jsin 45°) from Example 6.5.1 "" 1.414(cos ll' + j sin ll' ) = 1.414L ll' 4 4 4 .Jr
= 1,414e
J4
with the angle in radians.
Complex Numbers
74
[Ch.6
DO TIllS NOW! 6.4 1. You have already converted Z = 2 + j5 and w = -1 - j2 to polar form in "DO THIS NOW 6.3!". Convert z and w to exponential form.
2. Euler's relationship allows you to find more complicated complex exponentials. If z =x + jy, the result allows you to find jy x jy e' =e x+ =e e =eX(cosy+jsiny).
Evaluate the following (by hand and/or by technology - how directly will your machine do this? Find out NOW): (a)
ej
(b)
e-1.I+3j
(c)
Re(e-1.I+3 j )
(d)
Im(e-1.1+3j).
3. If you plot Re(e(-1.1+3j)x) you are plotting the graph of y
= e-l.1x cos(3x) .
Satisfy yourself about this both algebraically and by plotting it on your technology. 4. Experiment with curves of the type
y
=e
QX
cos(bx)
and y = e sin(bx) , as suggested below. QX
Describe in words what happens when a is (a) positive, (b) 0, (c) negative. Describe what happens when b is changed from small to large values. Why do you think this type of function is useful in engineering? What kind of motion can it represent? 5. If
z = -0.1 + j3
and t is an independent variable, use Euler's relationship to
write down (b)
Im(e
2zt)
Predict the shape of the graphs of both these functions for t ~ 0 and confirm by plotting them on your technology. Explain what you see in words. If you use your technology to turn symbols into pictures, you can "see" the importance and use of Euler's relationship in engineering. Figures 6.10(a) and (b) show an example of how a complex exponential can represent damped oscillations:
Figures 6.1O(a) and (b) Complex exponentials represent damped oscillations
Some uses of Polar and Exponential Form
Sec. 6.7]
75
6.7 SOME USES OF POLAR AND EXPONENTIAL FORM It turns out that it is particularly neat to handle multiplication, division and powers of complex numbers in polar form. Here is why. Have a look at these two complex numbers: ZI
'8
= rlLBI = 'ieJ
I
'(J
and Z2 = r 2LB2 = r 2e J 2.
Then their product is Gust using the laws of indices) ZIZ2
= ('iLBl)(r2LB2) = ('iej~ )(r2e
j(J2)
= 'ir2ej(~+(J2) = 'i r2L(BI +(2 ) .
I To multiply two complex numbers, multiply the moduli and add the arguments. Their quotient is (again just using the laws of indices)
.s., (rIL81) = (rle. 8) =.2-ej(~-82) =.2-L(8 j~
Z2
(r2 L B2)
(r2eJ 2)
r2
1
r2
-( 2).
I To divide two complex numbers, divide the moduli and subtract the arguments. To raise to a power n (once more the laws of indices are needed):
To raise a complex number to the power n, raise the modulus to the power n and multiply the argument by n, Example 6.7.1 Let z = 3 - 2j and w = 1- j. These give (check on your technology) in polar form c c 1r Z "" (3.606LO.588 ) and w "" (1.414L - 4" ) . Plotting these on an Argand diagram together with jz,
pz , j3z and z.w
gives Figure 6.11:
---
zw=(5.1<-.2D]
).).z
1·I·j·z
Figure 6.11 The picture shows the same story as the algebra
76
Complex Numbers
[Ch.6
The multiplication by powers of j again illustrates the rotational aspect of j (900
in the anticlockwise direction) and zw shows the addition of angles 0.588 + (- f) and multiplication of the moduli (3.606*1.414).
DO THESE NOW! 6.5 1. Convert the numbers 2 + 3j and 4 - 3j to polar form. HENCE evaluate the following: ..) (11
(i) (2+3j)(4-3j)
2+3'J 4-3j
(iii)
--
2
( J 2 + 3j 4-3j
Check your answers with your technology (Beware! Make sure you know and understand in what form they appear.) Show the two original numbers on an Argand diagram. Add to the diagram your answers to (i), (ii) and (iii), explaining how the rules in polar form, particularly concerning adding and subtracting angles, show themselves on the diagram. 2. Evaluate (a) (2+3j)3
(b)
(0.1-0.2j)9
Check your answers here by using technology.
6.8 COMPLEX ALGEBRA All of the above principles apply when you deal with complex algebra as opposed to just arithmetic, that is, with letters instead of or as well as numbers. The main differences are (a) you will have to do some algebra (!), and (b) you will need to use a CAS technology to check answers or to work out the more complicated cases. Example 6.8.1 Figures 6.12(a) and (b) show a few simple algebraic examples, carried out on handheld CAS technology, illustrating some basic rules. Can you identify them?
-liPrgPlIOlTeClean
J(
f l · III n · f~· •l:"11l!lf ~ Algebra Calc Other .~a + b'i)'~c + d'i)
f'·
Up,T
II:l~lAI ;~bralc~lclO{~;r IPr~Pl I 0 be~'; Up1
a'c - b·d +(a'd + b'c)'i • tCollect« rl .! al)'< r2 .! a2» -a',f2 a·,f2 . .,i '(al + a2)'rl'r2 2 T + ""2-'b. & .,i '(al - a2)'rl • tC 11 t« rl .! a1)) o ec
•
;x)
.<2.!45O) < 3 .!
-x)
ttll"'*"1ffJJ MAIN
fS
AD AurD
.,
fUNe )/99
'x
_'dliC"riti.Wil MAIN
UD Aura
fUNC )/'
Figures 6.12(a) and (b) Complex algebra on a hand-held CAS system
Sec. 6.8]
Complex Algebra
77
DO THIS NOW! 6.6 Find out NOW how your CAS handles complex numbers. Test it out on the examples below, which as usual you should do on paper as well as on your machine. I. Find the real and imaginary parts of z when I 2 1 -=--+-z 3+2j 4-3j 2. If z =a + jb and w = c + jd confirm by paper and pencil that your CAS is giving appropriate answers to the following: z+w
Izi
zw
z-w 3z-2w
zz
z w
1
arg(w)
(w+z)
You may also like to substitute some numerical values in for a, b, c and d for checking purposes. Make sure you know about and can use the different modes of CAS output (rectangular, polar or exponential forms) available to you. 6.9 ROOTS OF COMPLEX NUMBERS De Moivre'sTheorem
r
Note that (cos(B) + jsin(B))n = (e j 9 = e j n9 = cos(nB) + jsin(nB). This result is known as de Moivre's theorem. This helps to clarify and inform the way to raise a complex number to a fractional power - for instance to take square roots: zn
= r" e j n9 = r" (cos(nB) + j sin(nB)) .
In fact it is a little more complicated than this, since sine and cosine have period 21t, so increasing Bby 21t will change nothing. Thus in general h k'IS any In . t eger (0 ,_ +1,_+2,_, + 3....... ) ,and so z = re j9 = re j(9+2JZk) were
Normally the possibility of an extra 21t does not matter, but it does when n is a fraction, as it will indeed be when finding roots of complex numbers. Example 6.9.1 Solve the complex equation
z 2 = -9 .
In general exponential form
z2
=-9 =ge j(tr+2JZk)
Taking the square roots of both sides gives ..!.
z = (-9) 2
..!.
=9 2 e j (tr+2JZk )12 = 3e j (Jr+2JZk )/ 2 = 3e j (Jr /2+JZk)
Note that we take the positive square root of 9 here.
78
Complex Numbers
[Ch.6
To interpret this, remember that k can take any integer values. All values of z have modulus 3 - that is, they lie on the circle centred on the origin of radius 3. Now at what angles do they lie? Taking k
= 0 gives a z value at () =~ . Taking k = 1
gives a z value at () = 3~ . Taking further values of k simply repeats these two positions. There are thus just the two distinct values and just two square roots of -9. A diagram is a good check here since all the roots have the same modulus and in this case have an angle of 21t12 between their arguments.
fm 21
No_2p1l2 ~.ba
arc-.. •f . .t ....... a
1
2.2
Re
2
Figure 6.13 Roots are equally spaced around the circle ~lUlmple6.9.2
Solve the equation Z3
=0.1- 0.2j.
In exponential form we have 0.1- 0.2j given by
0.606962e j (0.36905+211k / 3)
i.e. 0.606962e j (O.36905)
,
= .223607 e j (1.10715 + 211k)
so all solutions are
with k = 0, 1 and 2.
0.606962e j (2.46345) , 0.606962e j (4.455784) •
Note that in this case the roots are all separated by 21t13 (120°) and effectively cut = 0.606962 into 3 equal sections. the circle
Izl
6.10 MINI CAS~ STUDY Complex numbers occur, for instance, in two important areas of engineering: electronic circuit theory and differential equation theory. Because you meet more about differential equations in Chapter 16 onwards, we choose to present here a little case study about an electronic circuit. An alternating current flowing in the circuit shown in Figure 6.14 is given by I = 10 sin(aJt) (OJ is a constant and t is time). The corresponding voltage drops depend on OJ and on the resistance R, capacitance C and inductance L of the circuit.
Sec. 6.10]
79
Complex Numbers: Exercises
For a resistance R, the voltage V is V = loR sin(ox) and this is "in phase" with the current. R L
c Figure 6.14 A simple series LCR circuit
For a capacitance C, the voltage V
=!JL sineOJ t c£
1! ) , a phase "lag" of
2
the current.
!!.2
behind
For an inductance L, the voltage V = OJ L 10 sin(OJt + 1!), a phase "lead" of
2
ahead of the current.
1r
2
If the current I = 10 sin(wt) is regarded as the imaginary part of a complex number
I
= Im(loe jOX ) then these various voltage drops can be represented very
succinctly
by recalling (refer back to Figure 6.11) that multiplication by j gives a lag of division by j (which is the same as multiplication by -j) gives a lead of Thus for resistance
V=RI.
For capacitance
V = -jI .
For inductance
V=jOJLI.
!!.2
and
!!... 2
«c
In general if V = IZ the factor Z is called the complex impedance, which is thus Z= R for a resistor, Z =
_--L for a capacitor, and ox:
Z = jOJL for an inductor.
END OF CHAPTER 6 - MIXED EXERCISES - DO ALL THESE NOW!
For each of the following, predict the answer by hand, check and observe by your technology, and explain and put right any discrepancies. 1. Let z\ = 2 - j, Zz = -1 + 2j, Z3 = 3 + 4j . Evaluate without using polar form, and present your answers in Cartesian form: (ii)
80
Complex Numbers (iii) z\ Z2
[Ch.6
(iv)
Z3
2. Where it is an advantage to do so, repeat the calculations of question 1 using polar form, and present your answers in polar form. 3. If Z =-o.1+5jand t is an independent variable, use Euler's relationship to write down and plot the resulting function of t, for t ~ 0 in each of (i) and (ii): (ii) Im(2e zt )
•
4. The voltage E across an alternating current circuit is obtained by multiplying the total impedance, Z, of the circuit by the current I flowing in the circuit: i.e,
E=I.Z.
A certain circuit, in which a current I = 4 + 5j flows, consists of three separate impedances, Z\, Z2 and Z3 arranged so that the total impedance Z is given by Z = Z\ + (Z2Z3) I(Z2 + Z3)' Obtain the symbolic expression for the voltage E and calculate its value (lEI), given (i)
Z\ = 2+3j
Z2 =4+3j
Z3 =2-5j
(ii)
Z\ = a+bj
Z2=4+3j
Z3 =2-5j.
5.
Solve the equation
6.
(Harder) Given -9+40j
Z3
= 8 + j8 and plot the results on an Argand diagram.
z = 4 + 5 j, write the following numbers in terms of z.
.09756-.12195j
-236-115j
1.7921c
.15617.